But the coefficients of γjsubscriptγj\gamma_{j} in the parentheses are elements of the ring ??\mathcal{O}, whence the last sum form of αα\alpha shows that α∈?α?\alpha\in\mathfrak{c}. Consequently, ?⊆???\mathfrak{a\subseteq c}.

Lemma 2. Any nonzero element αα\alpha of ??\mathcal{O} belongs only to a finitenumber of ideals of ??\mathcal{O}.

Proof. Let ?=(α1,…,αr)?subscriptα1normal-…subscriptαr\mathfrak{a}=(\alpha_{1},\ldots,\alpha_{r}) be any ideal containing αα\alpha and let
{ϱ1,…,ϱm}subscriptϱ1normal-…subscriptϱm\{\varrho_{1},\ldots,\varrho_{m}\} be a complete residue system modulo αα\alpha (cf. congruence in algebraic number field). Then

there can be different ideals ??\mathfrak{a} only a finite number, at most
1+m+(n2)+…+(mm)=2m1mbinomialn2normal-…binomialmmsuperscript2m1\!+\!m\!+\!{n\choose 2}\!+\ldots+\!{m\choose m}=2^{m}.

Lemma 3. Each ideal ??\mathfrak{a} of ??\mathcal{O} has only a finite number of ideal factors.

Proof. If ?∣?fragmentscnormal-∣a\mathfrak{c\mid a} and α∈?α?\alpha\in\mathfrak{a}, then by Lemma 1,
α∈?α?\alpha\in\mathfrak{c}, whence Lemma 2 implies that there is only a finite number of such factors ??\mathfrak{c}.

Lemma 4. All nonzero ideals of ??\mathcal{O} are cancellative, i.e. if ?⁢?=?⁢?????\mathfrak{ac=ad} then ?=???\mathfrak{c=d}.

If ?=(γ1,…,γs)?subscriptγ1normal-…subscriptγs\mathfrak{c}=(\gamma_{1},\ldots,\gamma_{s}) and ?=(δ1,…,δt)?subscriptδ1normal-…subscriptδt\mathfrak{d}=(\delta_{1},\ldots,\delta_{t}), we thus have the equation

Consequently, the generatorsγi=λi⁢1⁢δ1+…+λi⁢t⁢δtsubscriptγisubscriptλi1subscriptδ1normal-…subscriptλitsubscriptδt\gamma_{i}=\lambda_{{i1}}\delta_{1}+\ldots+\lambda_{{it}}\delta_{t} of ??\mathfrak{c} belong to the ideal ??\mathfrak{d}, and therefore ?⊆???\mathfrak{c\subseteq d}. Similarly one gets the reverse containment.

Lemma 5. If ?=?⁢????\mathfrak{a=bc} and ?≠(1)?1\mathfrak{b}\neq(1), then ??\mathfrak{c} has less ideal factors than ??\mathfrak{a}.

Proof. Evidently, any factor of ??\mathfrak{c} is a factor of ??\mathfrak{a}. But
?∣?fragmentsanormal-∣a\mathfrak{a\mid a} and ?∤?not-divides??\mathfrak{a\nmid c}, since otherwise we had
?=?⁢?=?⁢?⁢???????\mathfrak{c=ad=bcd} whence (1)=?⁢?1??(1)=\mathfrak{bd} which would, by Lemma 4, imply ?=(1)?1\mathfrak{b}=(1).

Lemma 6. Any proper ideal??\mathfrak{a} of ??\mathcal{O} has a prime ideal factor.

Proof. Let ??\mathfrak{c} be such a factor of ??\mathfrak{a} that has as few factors as possible. Then
??\mathfrak{c} must be a prime ideal, because otherwise we had ?=?1⁢??subscript?1?\mathfrak{c}=\mathfrak{c}_{1}\mathfrak{d} where
?1subscript?1\mathfrak{c}_{1} and ??\mathfrak{d} are proper ideals of ??\mathcal{O} and, by Lemma 5, the ideal ?1subscript?1\mathfrak{c}_{1} would have less factors than ??\mathfrak{c}; this however contradicts the fact ?1∣?fragmentssubscript?1normal-∣a\mathfrak{c}_{1}\mid\mathfrak{a}.

Lemma 7. Every nonzero proper ideal ??\mathfrak{a} of ??\mathcal{O} can be written as a product
?1⁢⋯⁢?ksubscript?1normal-⋯subscript?k\mathfrak{p}_{1}\cdots\mathfrak{p}_{k} where k>0k0k>0 and the factors ?isubscript?i\mathfrak{p}_{i} are prime ideals.

Proof. If ??\mathfrak{a} has only one factor ??\mathfrak{p} distinct from (1)1(1), then ?=???\mathfrak{a}=\mathfrak{p} is a prime ideal.
Induction hypothesis: Lemma 7 is in force always when ??\mathfrak{a} has at most nnn factors. Let ??\mathfrak{a} now have n+1n1n\!+\!1 factors. Lemma 6 implies that there is a prime ideal ??\mathfrak{p} such that
?=?⁢????\mathfrak{a=pd} where ?≠(1)?1\mathfrak{d}\neq(1) and ??\mathfrak{d} has, by Lemma 5, at most nnn factors. Hence, ?=?1⁢⋯⁢?k?subscript?1normal-⋯subscript?k\mathfrak{d}=\mathfrak{p}_{1}\cdots\mathfrak{p}_{k} and therefore,
?=?⁢?1⁢⋯⁢?k??subscript?1normal-⋯subscript?k\mathfrak{a}=\mathfrak{p}\mathfrak{p}_{1}\cdots\mathfrak{p}_{k} where all ??\mathfrak{p}’s are prime ideals.

of a nonzero ideal ??\mathfrak{a} of ??\mathcal{O} are identical, i.e. r=srsr=s and each prime factor ?isubscript?i\mathfrak{p}_{i} is equal to a prime factor ?jsubscript?j\mathfrak{q}_{j} and vice versa.

Proof. Any prime ideal has the property that if it divides a product of ideals, it divides one of the factors of the product; now these factors are prime ideals and therefore the prime ideal coincides with one of the factors. Similarly as in the proof of the fundamental theorem of arithmetics, one sees the uniqueness of the prime factorisation of ??\mathfrak{a}.